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Given $X, Y, Z$ i.i.d random variables the probability $P(X+Y>Z)$ can be found by the following three approaches:

  1. $X+Y-Z > 0$ region cuts the $X Y Z$ volume into two equal volumes as $X+Y-Z=0$ plane passes through origin. Thus, $P(X+Y>Z)=\frac{1}{2}$ which seems doubtful as $X+Y$ is mostly more than $Z$.
  2. $$P(X+Y>Z) = E(I_{X+Y>Z}) = E(E(I_{X+Y>Z}|Z))$$ $$E(E(I_{X+Y>Z}|Z)) = E(P(X+Y>z)) = E((1-F_{X+Y}(z))) = \int_z (1-F_{X+Y}(z))f_{Z}(z)dz$$ where, $$F_{X+Y}(z) = \int_{-\infty} ^{z} \int_{-\infty} ^{\infty} f_{X}(k)f_{Y}(h-k)dk$$

It is very complicated to substitute $F_{X+Y}(z)$ in the final expression of $P(X+Y>Z)$. Thus, it's quite difficult to comment on $P(X+Y>Z)$. Please help in this approach to solve it.

  1. $E[X] = E[Y] = E[Z]$, because $X, Y, Z$ follow identical, independent distributions.

$E[X + Y] = 2\times E[Z]$,

Intuition says that $P(X+Y>Z)$ thus ought to be 2/3, but I do not know if this is correct, nor how to make this reasoning rigorous.

Help in figuring out which approaches are correct, wrong and why ?

Wupadrasta Santosh
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  • If $X,Y,Z$ are i.i.d. continuous random variables with a distribution symmetric about $0$, then I find (1) convincing, but otherwise not, and particularly unconvincing for the reason you give if they are positive random variables. Meanwhile (3) is completely unconvincing in most plausible cases – Henry Oct 06 '17 at 11:26
  • The argument 1 is wrong in general. A hyperplane passing through the origin and cutting the whole space into two half-spaces does not mean that the probability in each half-space is equally likely. That is also a reason why you will learn many different distributions which assign different "weight" to different positions - they are not necessarily centered at origin and symmetrical as mentioned by Henry. – BGM Oct 06 '17 at 11:28
  • https://math.stackexchange.com/q/146450/321264 – StubbornAtom Jan 20 '21 at 09:47

2 Answers2

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Using Bayes theorem: \begin{align} P(X+Y>Z) & =\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy~f(x,y)P(X+Y>Z|X=x,Y=y)\\ &=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy~f(x)~f(y)P(Z<x+y)\\ &=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy~f(x)~f(y)\int_{-\infty}^{x+y}dz~f(z) \end{align}

Deep
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Let's take the following random variables: $X,Y,Z$ all take $0$ or $1$ with probability $\frac12$ and are independent.

Consider the possible cases which are of the same probability $\left(\frac18, \frac18, ...\right)$:

$$\begin{matrix} \color{white}{\checkmark}XYZ\\ \color{white}{\checkmark}\ 0\ 0\ 0\ \\ \color{white}{\checkmark}\ 0\ 0\ 1\\ \color{green}{\checkmark}0\ 1\ 0\\ \color{white}{\checkmark}0\ 1\ 1\\ \color{green}{\checkmark}1\ 0\ 0\\ \color{white}{\checkmark}1\ 0\ 1\\ \color{green}{\checkmark}1\ 1\ 0\\ \color{green}{\checkmark}1\ 1\ 1\end{matrix}$$

$X+Y>Z$ takes place where the check marks are. The probability of the event is $$\frac12$$

and not $$\frac23.$$

zoli
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